TrigonometryHardTrigonometric Equations ShareTrigonometric Equations · Hard problemSolve 2sin (x2)=12\sin\!\left(\dfrac{x}{2}\right)=12sin(2x)=1 on [0,2π)[0,2\pi)[0,2π).Ax=π3, 5π3x=\dfrac{\pi}{3},\ \dfrac{5\pi}{3}x=3π, 35πBx=π3, 5π4x=\dfrac{\pi}{3},\ \dfrac{5\pi}{4}x=3π, 45πCx=3π, 5π3x=\dfrac{3}{\pi},\ \dfrac{5\pi}{3}x=π3, 35πDx=π6, 5π3x=\dfrac{\pi}{6},\ \dfrac{5\pi}{3}x=6π, 35πCheck Show answer Walk through it step by step Practice more like this Review this concept